Big Ideas Math A Bridge To Success Algebra 2: Student Edition 2015

1st Edition

ISBN: 9781680331165

Author: HOUGHTON MIFFLIN HARCOURT

Publisher: Houghton Mifflin Harcourt

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Question

Chapter 2.1, Problem 45E

To determine

To Write: An equation of the form y=a(x−h)2+k with vertex (33,5) that models the flight path.

Expert Solution

Answer to Problem 45E

The equation of the form y=a(x−h)2+k with vertex (33,5) that models the flight path is,

y=−0.00459(x−33)2+5.

Explanation of Solution

Given: The fish leaves the water at (0,0)

Concept Used:

The equation of the form y=a(x−h)2+k represents a parabola with vertex (h,k) .

Calculation:

The vertex is (33,5) , therefore, h=33 and k=5 . Since the fish leaves the tank at (0,0) , x=0,y=0 . Substitute these values in y=a(x−h)2+k to obtain the value of a.

y=a(x−h)2+k0=a(0−33)2+50=1089a+5−1089a=5a=−51089≈−0.00459

Substitute the value of a, h=33 and k=5 in y=a(x−h)2+k to write the equation that represent the path of the fish.

y=−0.00459(x−33)2+5

Conclusion:

The equation of the form y=a(x−h)2+k with vertex (33,5) that models the flight path is,

y=−0.00459(x−33)2+5

To determine

To Write: The range and domain of the function. What do they represent these values in this model.

Expert Solution

Answer to Problem 45E

The range is [0,5] .

Explanation of Solution

Given: The model that represents the flight of the path is y=−0.00459(x−33)2+5 .

Concept Used:

Domain denotes the distance the fish covers horizontally. Range is the maximum height that the fish can reach.

Calculation:

Calculate the domain by finding the distance covered by the fish horizontally. The distance is calculated by the value of the x-intercepts.

y=−0.00459(x−33)2+5

Substitute y=0 to obtain the values of x,

y=−0.00459(x−33)2+50=−0.00459(x−33)2+50.00459(x−33)2=5(x−33)2=50.00459x−33=±50.00459x−33=±33x=±33+33=66,0

Hence, the domain of the function is [0,66] .

Calculate the range by finding the height covered by the fish. The fish starts from (0,0) and reaches the vertex (33,5) . Therefore, the minimum height is 0 and it reaches the maximum height of 5 as the vertex represents the highest point.

Hence, the range is [0,5] .

Conclusion:

The domain of the function is [0,66]and it denotes the distance the fish covers horizontally.Rangeis [0,5]and it denotes the maximum height that the fish can reach.

To determine

Does the value of a change when the flight has vertex (30,4) .

Expert Solution

Answer to Problem 45E

The value of a is almost same as it was before.

Explanation of Solution

Given: The fish leaves the water at (0,0) .

Concept Used:

The equation of the form y=a(x−h)2+k represents a parabola with vertex (h,k) .

Calculation:

The vertex is (30,4) , therefore, h=33 and k=5 . Since the fish leaves the tank at (0,0) , x=0,y=0 . Substitute these values in y=a(x−h)2+k to obtain the value of a.

y=a(x−h)2+k0=a(0−30)2+40=900a+4−900a=4a=−4900≈−0.0044

The value of a shows a negligible change.

Conclusion:

The value of a is almost same as it was before.

Chapter 2.1, Problem 44E

Chapter 2.1, Problem 46E

Chapter 2 Solutions

Big Ideas Math A Bridge To Success Algebra 2: Student Edition 2015

Ch. 2.1 - Prob. 1E Ch. 2.1 - Prob. 2E Ch. 2.1 - Prob. 3E Ch. 2.1 - Prob. 4E Ch. 2.1 - Prob. 5E Ch. 2.1 - Prob. 6E Ch. 2.1 - Prob. 7E Ch. 2.1 - Prob. 8E Ch. 2.1 - Prob. 9E Ch. 2.1 - Prob. 10E

Ch. 2.1 - Prob. 11E Ch. 2.1 - Prob. 12E Ch. 2.1 - Prob. 13E Ch. 2.1 - Prob. 14E Ch. 2.1 - Prob. 15E Ch. 2.1 - Prob. 16E Ch. 2.1 - Prob. 17E Ch. 2.1 - Prob. 18E Ch. 2.1 - Prob. 19E Ch. 2.1 - Prob. 20E Ch. 2.1 - Prob. 21E Ch. 2.1 - Prob. 22E Ch. 2.1 - Prob. 23E Ch. 2.1 - Prob. 24E Ch. 2.1 - Prob. 25E Ch. 2.1 - Prob. 26E Ch. 2.1 - Prob. 27E Ch. 2.1 - Prob. 28E Ch. 2.1 - Prob. 29E Ch. 2.1 - Prob. 30E Ch. 2.1 - Prob. 31E Ch. 2.1 - Prob. 32E Ch. 2.1 - Prob. 33E Ch. 2.1 - Prob. 34E Ch. 2.1 - Prob. 35E Ch. 2.1 - Prob. 36E Ch. 2.1 - Prob. 37E Ch. 2.1 - Prob. 38E Ch. 2.1 - Prob. 39E Ch. 2.1 - Prob. 40E Ch. 2.1 - Prob. 41E Ch. 2.1 - Prob. 42E Ch. 2.1 - Prob. 43E Ch. 2.1 - Prob. 44E Ch. 2.1 - Prob. 45E Ch. 2.1 - Prob. 46E Ch. 2.1 - Prob. 47E Ch. 2.1 - Prob. 48E Ch. 2.1 - Prob. 49E Ch. 2.1 - Prob. 50E Ch. 2.1 - Prob. 51E Ch. 2.1 - Prob. 52E Ch. 2.2 - Prob. 1E Ch. 2.2 - Prob. 2E Ch. 2.2 - Prob. 3E Ch. 2.2 - Prob. 4E Ch. 2.2 - Prob. 5E Ch. 2.2 - Prob. 6E Ch. 2.2 - Prob. 7E Ch. 2.2 - Prob. 8E Ch. 2.2 - 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Prob. 60E Ch. 2.4 - Prob. 1E Ch. 2.4 - Prob. 2E Ch. 2.4 - Prob. 3E Ch. 2.4 - Prob. 4E Ch. 2.4 - Prob. 5E Ch. 2.4 - Prob. 6E Ch. 2.4 - Prob. 7E Ch. 2.4 - Prob. 8E Ch. 2.4 - Prob. 9E Ch. 2.4 - Prob. 10E Ch. 2.4 - Prob. 11E Ch. 2.4 - Prob. 12E Ch. 2.4 - Prob. 13E Ch. 2.4 - Prob. 14E Ch. 2.4 - Prob. 15E Ch. 2.4 - Prob. 16E Ch. 2.4 - Prob. 17E Ch. 2.4 - Prob. 18E Ch. 2.4 - Prob. 19E Ch. 2.4 - Prob. 20E Ch. 2.4 - Prob. 21E Ch. 2.4 - Prob. 22E Ch. 2.4 - Prob. 23E Ch. 2.4 - Prob. 24E Ch. 2.4 - Prob. 25E Ch. 2.4 - Prob. 26E Ch. 2.4 - Prob. 27E Ch. 2.4 - Prob. 28E Ch. 2.4 - Prob. 29E Ch. 2.4 - Prob. 30E Ch. 2.4 - Prob. 31E Ch. 2.4 - Prob. 32E Ch. 2.4 - Prob. 33E Ch. 2.4 - Prob. 34E Ch. 2.4 - Prob. 35E Ch. 2.4 - Prob. 36E Ch. 2.4 - Prob. 37E Ch. 2.4 - Prob. 38E Ch. 2.4 - Prob. 39E Ch. 2.4 - Prob. 40E Ch. 2.4 - Prob. 41E Ch. 2 - Prob. 1CR Ch. 2 - Prob. 2CR Ch. 2 - Prob. 3CR Ch. 2 - Prob. 4CR Ch. 2 - Prob. 5CR Ch. 2 - Prob. 6CR Ch. 2 - Prob. 7CR Ch. 2 - Prob. 8CR Ch. 2 - Prob. 9CR Ch. 2 - Prob. 10CR Ch. 2 - 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